you have ∆y ≈ k'(-2)*∆x
You have k(-2), so use ∆x=0.6
If the curve is concave down, it lies below the tangent line.
Suppose k is a continuous function so that k(-2)=3 and k'(-2)= -4 .
a. Use the given information to approximate the value of k(-1.4).
b. If k''(x)<0 for -3<=x<=0 , will the approximation for k(-1.4) you found in part (a) be greater than or less than the actual value of k(-1.4) ? Justify your answer.
c. Suppose that -5<=k''(x)<=-2 for -3<=x<=0 . Find the largest number L and the smallest number U so that L<=k'(0)<=U . Justify your answer.
3 answers
Sorry my question was solely on 5c as that is the only one I don't understand.
suppose k" = -5 on [-3,0]. You know that k'(-2) = -4. So, k'(0) would be -4+2(-5) = -14
Now, suppose that k" = -2 on [-3,0]. k' would be -4+2(-2) = -8
So, -14 <= k'(0) <= -8
remember that k" is just the slope of k'. k'(0) will lie between the steepest and shallowest tangent lines to k'
Now, suppose that k" = -2 on [-3,0]. k' would be -4+2(-2) = -8
So, -14 <= k'(0) <= -8
remember that k" is just the slope of k'. k'(0) will lie between the steepest and shallowest tangent lines to k'
1 answer